Body movement question v(t), a(t), x(t)

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SUMMARY

The discussion focuses on the motion of a body in one dimension with a velocity function defined as v(t) = t^3 - 10t^2 + 16t for t ≥ 0. The initial speed is determined to be 0 m/s at t=0, while the initial acceleration is calculated as 16 m/s². The position function x(t) is derived as x(t) = (1/4)t^4 - (10/3)t^3 + 8t², and the times when the body passes through the origin are found to be t=0s, t=3.139s, and t=10.194s. To find the extreme values of distance from the origin, it is suggested to analyze when the velocity equals zero.

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I don't learn in English so I may phrase words incorrectly. Just let me know if something doesn't make sense

Homework Statement



A body moves in one dimension where his speed is dependent on time:
[tex]v(t) = t^3-10t^2+16t, t \geq0[/tex]. v is in m/s amd t is in s.

a) what is it's initial speed?
b) what is it's initial acceleration?
c) find x(t) if it's known that x(0)=0
d) at what time does the body pass through the origin (x=0)?
e) find the extreme values (max and min) of the distance between the body and the origin, from the beginning of its movement and until the last time it passes through the origin

The Attempt at a Solution



a) initial speed is at t=0, so v=0 m/s
b) initial acceleration is at t=0 [tex]\rightarrow a(t) = 3t^2-20t+16 \rightarrow t=0, a=16 m/s^2[/tex]
c) [tex]x(t) = \frac{1}{4}t^4 - \frac{10}{3}t^3 + 8t^2[/tex] ------not sure about (c)
d) [tex]\frac{1}{4}t^4 - \frac{10}{3}t^3 + 8t^2=0 \rightarrow t=0s, t=10.194s, t=3.139s[/tex]
e) I am not sure about this one: wouldn't the min distance be 0m. since if x=0, there is no distance. and what do I do for the max distance? how do I know what the max distance is?

Thanks!
 
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You can find the distance extrema by looking for when velocity=0. Everything else looks good.
 

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